2015 AIME I Problems/Problem 4
Point lies on line segment with and . Points and lie on the same side of line forming equilateral triangles and . Let be the midpoint of , and be the midpoint of . The area of is . Find .
Diagram by RedFireTruck (talk) 18:52, 15 February 2021 (EST)
Let be the origin, so and Using equilateral triangle properties tells us that and as well. Therefore, and Applying the Shoelace Theorem to triangle gives
Note that and . Also, . Thus, by SAS.
From this, it is clear that a rotation about will map to . This rotation also maps to . Thus, and . Thus, is equilateral.
Using the Law of Cosines on , Thus, .
Using Stewart's Theorem on ,
Calculating the area of , Thus, , so . Our final answer is .
Admittedly, this is much more tedious than the coordinate solutions.
I also noticed that there are two more ways of showing that is equilateral:
One way is to show that , , and are related by a spiral similarity centered at .
The other way is to use the Mean Geometry Theorem. Note that and are similar and have the same orientation. Note that is the weighted average of and , is the weighted average of and , and is the weighted average of and . The weights are the same for all three averages. (The weights are actually just and , so these are also unweighted averages.) Thus, by the Mean Geometry Theorem, is similar to both and , which means that is equilateral.
Note: A much easier way to go about finding without having to use Stewart's Theorem is to simply drop the altitudes from M and E to AC, thus hitting AC at points X and Y. Then clearly AEY and AMX are similar with ratio 2. But we know that . Additionally, from similar triangles meaning we can now just do pythagorean theorem on right triangle to get - SuperJJ
Medians are equal, so
is equilateral triangle.
The height of is distance from to midpoint is
is the median of
The area of
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